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3 Measure and Integration Then f n is a simple function and positive whenever f is positive (the latter follows from the fact that in this case f −1 [ Bn,k ] = ∅ for all k, n). Moreover, f n ( x ) → f ( x ). Indeed, if x ∈ Ω, then there exists an n0 ∈ N such that | f ( x )| ≤ 2n0 . Then | f n ( x ) − f ( x )| ≤ 2−n for all n ≥ n0 . If f ≥ 0 then f n ≤ f n+1 . Indeed, if f n ( x ) = k2−n then f ( x ) ∈ [k2−n , (k + 1)2−n ). But then either f ( x ) ∈ [(2k)2−(n+1) , (2k + 1)2−(n+1) ), in which case f n+1 ( x ) = (2k)2−(n+1) = f n ( x ), or f ( x ) ∈ [(2k + 1)2−(n+1) , (2k + 2)2−(n+1) ), in which case f n+1 ( x ) = (2k + 1)2−(n+1) > f n ( x ). In the case where K = C, we find sequences of simple functions ( gn ) and (hn ) that converge to Re f and Im f , respectively. Then we set f n := gn + ihn for all n ∈ N. For example, this is natural for stopping times of random processes. There exist metrics on R resp. [0, ∞] such that B (R) resp. B ([0, ∞]) is the Borel σ-algebra for this metric. We sometimes prefer to work with a slightly more general notion of measurable functions. It has technical advantages to allow functions to take the values ∞ or −∞. Remark 3.54 (Extended real line). We put R := {−∞} ∪ R ∪ {∞} which we endow with the σ-algebra B (R), defined as σ(B (R) ∪ {{−∞}, {∞}}). It follows that a function f : (Ω, Σ) → R is measurable if and only if f −1 [{∞}], f −1 [{−∞}] ∈ Σ and f −1 [ A] ∈ Σ for all A ∈ B (R). Similarly, B ([0, ∞]) is defined as σ (B ([0, ∞)) ∪ {{∞}}). We also remark that Proposition 3.53 generalizes to this situation. Exercise 3.55. Let (Ω, Σ) be a measurable space and f n : Ω → R be measurable for all n ∈ N. Then the functions lim infn→∞ f n , lim supn→∞ f n , infn∈N f n and supn∈N f n are measurable. Hint: Observe that { x ∈ Ω : sup{ f n ( x ) : n ∈ N} > a} = S n ∈N { x ∈ Ω : f n ( x ) > a }. 3.7 The Lebesgue integral Given a measure space (Ω, Σ, µ), we now introduce the Lebesgue integral R f dµ for suitable complex-valued, measurable functions that we call ‘integrable’. We proceed in several steps and first define the integral for (real-valued) measurable functions f taking values in [0, ∞]. Definition 3.56. Let (Ω, Σ, µ) be a measure space. If f : Ω → [0, ∞] is a simple function, with standard representation f = ∑nk=1 ak 1 Ak , one defines the Lebesgue integral of f by Z n f dµ := ∑ ak µ( Ak ) k =1 72 3.7 The Lebesgue integral where, by convention, 0 · ∞ := 0 and ∞ + ∞ = ∞. The following lemma shows that the Lebesgue integral for nonnegative simple functions is positively linear and monotone. Lemma 3.57. If (Ω, Σ, µ) is R a measure space,R f , g : Ω R→ [0, ∞] are simple functions and λ > 0, then λ f + g dµ = λ f dµ + g dµ. Moreover, if R R f ≤ g, then f dµ ≤ g dµ. n Proof. If ∑nk=1 ak 1 Ak is the standard representation R of f , then R ∑k=1 λak 1 Ak is the standard representation of λ f . Now λ f dµ = λ f dµ follows from the definition of the integral. Now let λ = 1 and ∑m l =1 bl 1 Bl be the w standard representation of g. If ∑ j=1 c j 1Cj is the standard representation S of f + g, then Cj = ak +bl =c j Ak ∩ Bl . Thus Z f dµ + Z n g dµ = ∑ k =1 w = m ak µ( Ak ) + ∑ bl µ( Bl ) = l =1 ∑ c j µ(Cj ) = Z n m ∑ ∑ (ak + bl )µ( Ak ∩ Bl ) k =1 l =1 f + g dµ, j =1 where we have used the finite additivity of µ and the fact that the sets Ak ∩ Bl are pairwise disjoint. Now assume that f ≤ g. Consider the function g − f , with the convention that ∞ − ∞ = 0 and ∞ − c = ∞ for all c ∈ [0, ∞). Then g − f is a nonnegative, simple function. Hence, Z g dµ = Z ( g − f ) + f dµ = Z g − f dµ + Z f dµ ≥ Z f dµ, since the integral of a nonnegative, simple function is clearly nonnegative. Lemma 3.58. Let (Ω, Σ, µ) be a measure R space and f : Ω → [0, ∞] be simple. Then ν : Σ → [0, ∞] defined by ν( A) = 1 A f dµ is a measure on (Ω, Σ). Proof. As 1∅ f is the zero function, the map ν clearly satisfies property (M1). It remains to very (M2). So let ( Ak ) be a sequence of disjoint sets in S Σ. Define Bn := nk=1 Ak . Then ν( Bn ) = Z 1Bn f dµ = Z n ∑ k =1 1 Ak f dµ = n Z ∑ n 1 Ak f dµ = ∑ ν ( A k ). k =1 k =1 Clearly, Bn ⊂ Bn+1 for all n ∈ N and A := n∈N Bn = k∈N Ak . Since f is a simple function, take f = ∑m j=1 a j 1Cj to be its standard representation. S S 73 The convention 0 · ∞ = 0 ensures that the Lebesgue integral of the zero function on R is zero. Note that λ(R) = ∞. 3 Measure and Integration Observe that ν( A) = Z 1 A f dµ = Z m ∑ a j 1Cj ∩ A dµ j =1 m = ∑ a j µ(Cj ∩ A) j =1 m ∑ a j µ(Cj ∩ Bn ) n→∞ = lim j =1 = lim n→∞ = lim m Z ∑ a j 1Cj ∩Bn dµ j =1 Z n→∞ ∞ 1Bn f dµ = lim ν( Bn ) = n→∞ ∑ ν( Ak ) k =1 In the previous calculation we used the definition of the integral for simple functions, that it is positively linear, the continuity of the measure µ, and in the last step the identity ν( Bn ) = ∑nk=1 ν( Ak ) that was established above. This proves (M2). Hence ν is a measure. We can now define the Lebesgue integral by approximation for an arbitrary measurable function f : Ω → [0, ∞]. Definition 3.59. Let (Ω, Σ, µ) be a measure space and f : Ω → [0, ∞] measurable. One defines Z nZ o : f dµ = sup g dµ 0 ≤ g ≤ f , g simple . We say that f is (Lebesgue) integrable, if R f dµ < ∞. Example 3.60. If δa is the Dirac measure on (Ω, Σ) for an a ∈ Ω, then for all measurable f : Ω → [0, ∞] we have Z f dδa = f ( a) and f is integrable if and only if f ( a) < ∞. Proof. First, let f be a simple function. Then δa ( f −1 [{t}]) = R1 if a ∈ f −1 [{t}], i.e. f ( a) = t and δa [ f −1 [{t}]] = 0 else. Thus in this case f dδa = f ( a). Now let f : Ω → [0, ∞] be a measurable function. If g is a simple function with g ≤ f , then Z Ω g dδa = g( a) ≤ f ( a). Taking the supremum over such g, it follows that 74 R f dµ ≤ f ( a). 3.7 The Lebesgue integral For the reverse inequality, let g( x ) := f ( a)1 f −1 [{ f (a)}] ( x ). Since f is measurable and { f ( a)} ∈ B (R), it follows Rthat f −1 [{Rf ( a)}] ∈ Σ. So g is a simple function. Moreover, g ≤ f . Hence, f dδa ≥ g dδa = f ( a). Exercise 3.61. Consider the measure space (N, P (N), ζ ). A measurable function f : N → [0, ∞] is a sequence ( an )n∈N = ( f (n))n∈N in [0, ∞]. Show that f is integrable if and only if ( an ) ∈ `1 and in this case, ∞ Z N f dζ = ∑ an . n =1 Theorem 3.62 (Monotone convergence theorem). Let (Ω, Σ, µ) be a measure space and f n : Ω → [0, ∞] be measurable functions for all n ∈ N that are monotonically increasing in n. Then f ( x ) := supn∈N f n ( x ) is measurable and R R f dµ = supn∈N f n dµ. Proof. As a consequence of Exercise 3.55, f is measurable. One deduces from f n ≤ f that sup n ∈N Z f n dµ ≤ Z f dµ. It remains to show the converse inequality. To this end, let 0 ≤ ϕ ≤ f be simple and β > 1. It suffices to show that Z ϕ dµ ≤ β sup n ∈N Z f n dµ. The sufficiency follows from first taking the supremum over all simple 0 ≤ ϕ ≤ f and then taking the infimum over all β > 1. Define Bn := S { x ∈ Ω : β f n ( x ) ≥ ϕ( x )}. Then Bn ∈ Σ, Bn ⊂ Bn+1 andR n∈N Bn = Ω. Moreover, we have β f n ≥ ϕ on Bn . Define ν( A) := 1 A ϕ dµ for all A ∈ Σ. Then ν is a measure by Lemma 3.58. Hence Z ϕ dµ = ν(Ω) = lim ν( Bn ) n→∞ = lim Z ϕ1Bn dµ n→∞ ≤ sup Z n ∈N = β sup n ∈N β f n dµ Z f n dµ. This establishes the converse inequality. Hence supn∈N R f n dµ = R f dµ. Remark 3.63. Let f : Ω → [0, ∞] be measurable. Let ( f n ) be a sequence of nonnegative measurable functions such that f n ↑ f pointwise. Note 75 Instead of sup we could write lim in the statement of Theorem 3.62. Of course this requires to consider improper limits, thereby for example setting f ( x ) to ∞ where ( f n ( x )) is unbounded. 3 Measure and Integration that such a sequence exists by Proposition 3.53. It now follows from Theorem 3.62 that Z Z nZ o f dµ = sup ϕ dµ : 0 ≤ ϕ ≤ f simple = sup f n dµ. n ∈N So while the integral for nonnegative measurable functions is defined as the supremum over a very large class of simple functions, it is actually already obtained by the supremum over an approximating sequence as in Proposition 3.53. Corollary 3.64. Let space, R (Ω, Σ, µ) be a measure R R f , g : Ω → [0, ∞] measurable and R λ > 0.R Then λ f + g dµ = λ f dµ + g dµ. Moreover, if f ≤ g, then f dµ ≤ g dµ. Proof. Let ( f n ) and ( gn ) be sequences of simple functions such that f n ↑ f and gn ↑ g. Then λ f n + gn ↑ λ f + g. Using monotone convergence and Lemma 3.57, we obtain Z λ f + g dµ = lim Z n→∞ = lim λ n→∞ λ f n + gn dµ Z f n dµ + Z gn dµ = λ Z f dµ + Z g dµ The second assertion is clear from the definition and was already observed in the proof of Theorem 3.62. Note that the notion of a null set crucially depends on the measure, of course. So sometimes it is necessary to point out this dependence by speaking about a µ-null set. Definition 3.65. Let (Ω, Σ, µ) be a measure space. A null set is a set N ⊂ Ω such that there exists a set M ∈ Σ with N ⊂ M and µ( M ) = 0. Now let P = P( x ) be a property which, depending on x, may be true or false. We say that P holds almost everywhere or for almost every x ∈ Ω, if { x ∈ Ω : P( x ) is false} is a null set. If µ is a probability measure, we also say that P holds almost surely if it holds almost everywhere. Remark 3.66. Note that we do not assume that a null set N is measurable. A measure space in which all null sets are measurable is called complete. It is not hard to see that the measure space (Ω, M , µ∗ ) in the proof of Carathéodory’s theorem, Theorem 3.42, is always complete. We note that there is a straightforward procedure to complete a measure space by minimally enlarging the σ-algebra and thus the domain of the measure to make all null sets measurable. Example 3.67. Consider (R, B (R), δ0 ). Then almost every (or more precisely, δ0 -a.e.) x ∈ R is equal to 0. Consider (R, B (R), λ). Then almost every (or more precisely, λ-a.e.) x ∈ R is not equal to 0. In fact, almost every x ∈ R is irrational. Corollary 3.68. R Let (Ω, Σ, µ) be a measure space and f : Ω → [0, ∞] be measurable. Then f dµ = 0 if and only if f ( x ) = 0 for almost every x ∈ Ω. 76 3.7 The Lebesgue integral Proof. If f is a simple function, then the assertion R is obvious. In the general case, let a measurable f : Ω → [0, ∞] with f dµ = 0 be given. Let ( f n ) be an increasing sequence of simple functions converging to f . Such R a sequence exists by Proposition 3.53. It follows that f n dµ = 0 for all n ∈ N. Hence, by the case above, { x ∈ Ω : f n ( x ) 6= 0} is a null set, whence there exists Mn ∈ Σ with µ( Mn ) = 0 such that f n ( x ) = 0 for all S x 6∈ Mn . Put M := n∈N Mn . Then M ∈ Σ and µ( M) ≤ ∑∞ n=1 µ ( Mn ) = 0. Moreover, x 6∈ M implies that f ( x ) = 0. Hence f = 0 almost everywhere. If, conversely, f = 0 almost everywhere, then there exists a measurable set M with f ( x ) = 0 for all x 6∈ M. If g is a simple function with 0 ≤ g ≤ f then g−1 [{ x }] ⊂ R M for all x > 0. By the definition of the integral for R simple functions, g dµ = 0 and hence, since g ≤ f was arbitrary, f dµ = 0. Exercise 3.69. Let (Ω, Σ, µ) be a measure space and f : Ω → [0, ∞] be measurable. Show that ν : Σ → [0, ∞], defined by ν( A) := Z A f dµ := Z 1 A f dµ, defines a measure on (Ω, Σ). Moreover, show that if µ( A) = 0, then ν( A) = 0. In other words, ν is absolutely continuous with respect to µ, which is usually denoted by writing ν µ. Remark 3.70. In the setting of Example 3.69 the function f is called the density of ν with respect to µ. In probability theory a density for a distribution is commonly taken with respect to the Lebesgue measure. According to the Radon–Nikodym theorem a probability measure P on B (R) has a density with respect to the Lebesgue measure if and only if it is absolutely continuous with respect to the Lebesgue measure. Equivalently, this is the case if and only if its distribution function F ( x ) := P((−∞, x ]) is an absolutely continuous function. Not every continuous real function is absolutely continuous, but every Lipschitz function is. Theorem 3.71 (Fatou’s Lemma). Let (Ω, Σ, µ) be a measure space, f n : Ω → [0, ∞] be measurable and set f ( x ) := lim infn→∞ f n ( x ). Then f is measurable and Z Z f dµ ≤ lim inf f n dµ. n→∞ Proof. Note that f ( x ) = supk≥1 infn≥k f n ( x ). Let gk ( x ) := infn≥k f n ( x ). Then gk is measurable by Exercise 3.55. R Clearly gkR↑ f . So it follows from monotone convergence that supk∈N gk dµ = Ω f dµ. On the other hand, Z Z Z f dµ = sup gk dµ ≤ lim inf f n dµ. k ∈N n→∞ 77 3 Measure and Integration R R Indeed, for every n ≥ k we have gk ≤ f nRand thus gk dµR ≤ f n dµ. Since this is true for all n ≥ k, we have gk dµ ≤ infn≥k f n dµ. By taking the supremum over k ∈ N on both sides, the above inequality follows. Definition 3.72. Let (Ω, Σ, µ) be a measure R space, f : Ω → K be measurable. Then f is called integrable if | f | dµ < ∞. We write f ∈ L 1 (Ω, Σ, µ). If K = R, we set Z f dµ := Z + f dµ − Z f − dµ. Note that if | f | is integrable then f + and f − are both integrable nonnegative functions. If K = C, we set Z f dµ = Z Re f dµ + i Z Im f dµ. Note that if | f | is integrable, then Re f and Im f are both integrable realvalued functions. If f is an integrable (real or complex) measurable function and A ∈ Σ, we define Z Z f dµ := f 1 A dµ A Lemma 3.73. Let (Ω, Σ, µ) be a measure space. R R (a) For all integrable f , we have | f dµ| ≤ | f | dµ. R (b) If Rf is integrable and λ ∈ K, then λ f is integrable and λ f dµ = λ f dµ. R (c) If f and g are integrable, then f + g is integrable and f + g dµ = R R f dµ + g dµ. Remark 3.74. Note that (b) and (c) can be expressed by saying R that the integrable functions form a vector space and the map f 7→ f dµ is a linear map from the integrable functions to K. Proof. Let us first consider the case where K = R. (a) We have Z Z Z Z Z Z + − + − f dµ + f dµ = | f | dµ, f dµ = f dµ − f dµ ≤ where we have used Corollary 3.64 in the last step. R R (b) First note that by Corollary 3.64 one has |λ f | dµ = |λ| | f | dµ < ∞ if f is integrable. This proves that λ f is integrable whenever f is. 78 3.7 The Lebesgue integral Now, if λ > 0, then (λ f )+ = λ f + and (λ f )− = λ f − . Thus, using Corollary 3.64, Z λ f dµ = Z =λ Z + λ f dµ − Z + f dµ − λ λ f − dµ Z − f dµ = λ Z f dµ. If, on the other hand, λ < 0, then (λ f )+ = −λ f − and (λ f )− = −λ f + . Thus, in this case, Z λ f dµ = Z =λ − −λ f dµ − Z + f dµ − λ Z −λ f + dµ Z − f dµ = λ Z f dµ. (c) Since | f + g| ≤ | f | + | g|, it follows that Z | f + g| dµ ≤ Z | f | + | g| dµ = Z | f | dµ + Z | g| dµ < ∞ if f and g are integrable. Moreover, by definition and Corollary 3.64, Z f dµ + Z g dµ = = (∗) = = Z + f dµ − Z Z f + Z − f dµ + + + g dµ − + + g dµ − Z ( f − + g− ) dµ Z ( f + g)− dµ ( f + g) dµ − Z Z Z g− dµ f + g dµ. Here, (∗) follows from integrating the identity f + + g+ + ( f + g)− = ( f + g)+ + f − + g− and using Corollary 3.64. R In the case where K = C, for (b) and (c) one uses that f dµ = R R Re f dµ + i Im f dµ. We omit the easy computations. For (a), we use that for every complex number z one has |z| = supt∈R Re(eit z). So for every t ∈ R one has Re e it Z f dµ = Re Z it e f dµ = Z it Re(e f ) dµ ≤ Z | f | dµ. Taking the supremum over t ∈ R, statement (a) follows. Theorem 3.75 (Dominated convergence theorem). Let (Ω, Σ, µ) be a measure space and ( f n ) be a sequence of integrable functions with the following two properties. (a) f˜( x ) := limn→∞ f n ( x ) exists for almost every x ∈ Ω, say outside the set N ∈ Σ with µ( N ) = 0. 79 3 Measure and Integration The technicalities with f and f˜ are required as f˜ might not be measurable. Alternatively, to avoid speaking about f˜ and N, one could assume that a measurable f : Ω → K is given that is pointwise almost everywhere the limit of the f n . (b) There exists an integrable function g with | f n ( x )| ≤ g( x ) for almost every x ∈ Ω and all n ∈ N. Then f : Ω → K, defined by f ( x ) = f˜( x ) if x 6∈ N and f ( x ) = 0 if x ∈ N, is integrable and lim Z n→∞ | f n − f | dµ = 0. In particular, lim Z n→∞ f n dµ = Z f dµ. Proof. Changing f n and f on a set of measure zero, we may assume that (a) and (b) hold everywhere. By Proposition 3.30, f is measurable. Since | f | ≤ g, it follows that f is integrable. Now observe that | f n − f | ≤ 2g and hence 2g − | f n − f | ≥ 0. By Fatou’s Lemma 3.71, Z 2g dµ = = Z lim inf(2g − | f n − f |) dµ ≤ lim inf n→∞ Z n→∞ 2g dµ − lim sup n→∞ Z Z 2g − | f n − f | dµ | f n − f | dµ. R R So lim supn→∞ | f n − f | dµ = 0, and therefore limn→∞ | f n − f | dµ = 0. By Lemma 3.73, Z Z Z f n dµ − f dµ ≤ | f n − f | dµ → 0. This proves the claim. Example 3.76. Let us give an example that condition (b) in Theorem 3.75 is necessary. Consider (R, B (R), λ). If we set f n := n1(0, 1 ) , then ( f n ) is n aR sequence of simpleR functions converging to 0 everywhere. However, R f n dλ ≡ 1 6 → 0 = R 0 dλ. In the lecture we discussed here how the expected value of random variables is usually computed using densities for the push-forward measure. Recall that µΦ ( A) = µ(Φ−1 [ A]) for all A ∈ F , see Lemma 3.31. Exercise 3.77. Consider the situation of Exercise 3.69, i.e.R (Ω, Σ, µ) is a measure space, f : Ω → [0, ∞] is measurable and ν( A) := A f dµ. Show that g is integrable with respect to ν if and only if g f is integrable with respect to µ and in this case, Z g dν = Z g f dµ. We close this section by considering the integration under a push-forward measure, which is of great importance for applications. Theorem 3.78. Let (Ω, Σ, µ) be a measure space, ( M, F ) be a measurable space and Φ : (Ω, Σ) → ( M, F ) be measurable. We denote the push-forward of µ under Φ by µΦ . Then for a measurable f : ( M, F ) → (K, B (K)) we have 80 3.8 On the connection between the Lebesgue and Riemann integral f ◦ Φ ∈ L 1 (Ω, Σ, µ) if and only if f ∈ L 1 ( M, F , µΦ ). In this case, Z Ω Z f ◦ Φ dµ = M f dµΦ . Proof. First, let f = ∑nj=1 ak 1 Ak be a nonnegative, simple function. Then f ◦Φ = n ∑ a k 1 Φ −1 [ A k ] j =1 and thus, by the definition of the push-forward measure, Z Ω f ◦ Φ dµ = n ∑ ak µ(Φ −1 n [ A]) = j =1 ∑ ak µΦ ( A) = Z j =1 M f dµΦ . It follows that for nonnegative, simple functions the assertion holds true. Now let f : M → [0, ∞] be measurable and ( f n ) be a sequence of simple functions with f n ↑ f pointwise. Then, by monotone convergence and the above, Z Ω f ◦ Φ dµ = sup Z n ∈N Ω f n ◦ Φ dµ = sup Z n ∈N M f n dµΦ = Z M f dµΦ . This shows that the assertion holds for arbitrary measurable positive f . Since | f ◦ Φ| = | f | ◦ Φ, it follows that f ∈ L 1 ( M, F , µΦ ) if any only if f ◦ Φ ∈ L 1 (Ω, Σ, µ). The general formula follows by splitting real valued functions f into the positive functions f + and f − and complex valued functions f into Re f and Im f . 3.8 On the connection between the Lebesgue and Riemann integral We next compare the Lebesgue integral with the Riemann integral. As is well-known, every continuous function f : [ a, b] → R is Riemann integrable. We now show that such functions are also Lebesgue integrable and the Lebesgue integral agrees with the Riemann integral. Theorem 3.79. If f : [ a, b] → K is continuous, then f is a Lebesgue integrable function on ([ a, b], B ([ a, b]), λ). Moreover, Z b Z [ a,b] f dλ = R- a f (t) dt. Proof. Let a sequence of partitions πn := (t0(n) , . . . , t(knn ) ) with |πn | → 0 be given and let ξ n = (ξ 1(n) , . . . , ξ k(nn ) ) be a sequence of associated sample 81 3 Measure and Integration points. Put kn f n := ∑ f (ξ (j n) )1[t(j−n)1,t(j n) ) . j =1 R Then f n is a simple function and [a,b] f n dλ = S( f , πn , ξ n ). Moreover, (a) | f n | ≤ k f k∞ and the latter is integrable on our measure space and (b) f n (t) → f (t) for all t ∈ [ a, b]. Indeed, for fixed t ∈ [ a, b], we have | f n (t) − f (t)| = | f (ξ (jnn) ) − f (t)|, where ξ (jnn) is the sample point in the interval [t(jnn−) 1 , t(jnn) ] and jn is chosen such that t lies in this interval. But then |ξ (jnn) − t| ≤ |t(jnn) − t(jnn−) 1 | ≤ |πn | → 0 and hence, by the continuity of f , it follows that | f (ξ (jnn) ) − f (t)| → 0. Hence the dominated convergence theorem, Theorem 3.75, applies and shows that f is integrable and Z [ a,b] f dλ = lim Z n→∞ [ a,b] f n dλ = lim S( f , πn , ξ n ). n→∞ Rb Since, on the other hand, the Riemann sums converge to R- a f (t) dt, the assertion follows. Remark 3.80. Actually, the continuity assumption in Theorem 3.79 is not needed, if one is willing to enlarge the σ-algebra. It can be proved that if f : [ a, b] → K is Riemann integrable then it is almost everywhere equal to a measurable function that is Lebesgue integrable and the Riemann and the Lebesgue integral coincide. There is also an extension of Theorem 3.79 to improper Riemann integrals. We recall that if −∞ < a < b ≤ ∞ and f : [ a, b) → R is continuous, then R rf is called improperly Riemann integrable on [ a, b) if the limit limr↑b R- a f (t) dt exists. The limit is then called the improper Riemann R b− integral of f over [ a, b) and denoted by R- a f (t) dt. Theorem 3.81. Let −∞ < a < b ≤ ∞ and f : [ a, b) → R be continuous, R b− such that the improper Riemann integral R- a | f (t)| dt exists, then f is inR b− tegrable on ([ a, b), B ([ a, b)), λ), the improper Riemann integral R- a f (t)dt exists and Z Z b− [ a,b) f dλ = R- a f (t) dt. Proof. Pick a sequence (bn ) ⊂ ( a, b) with bn ↑ b. By monotone convergence and Theorem 3.79, Z [ a,b) | f | dλ = lim Z = lim Rn→∞ 82 | f | dλ n→∞ [ a,bn ] Z bn a Z b− | f (t)| dt = R- a | f (t)| dt < ∞. 3.8 On the connection between the Lebesgue and Riemann integral It follows that | f | is integrable on [ a, b). Moreover, since 1[a,bn ) f converges to f pointwise and |1[a,bn ) f | ≤ | f |, the dominated convergence theorem yields Z [ a,b) f dλ = lim Z bn Z n→∞ [ a,bn ) f dλ = lim Rn→∞ a Z b− f (t) dt = R- a f (t) dt, where we have used Theorem 3.79 in the second step. Remark 3.82. Similar results as in Theorem 3.81 also hold for improper Riemann integrals that are improper on the left-hand side or on both sides. R Example 3.83. In Theorem 3.81, the assumption that R- [a,b) | f (t)| dt exists is crucial and cannot be omitted. An example is given by f : [1, ∞) → R , defined by f (t) = sint t In this case, by integration by parts, we obtain Z x sin t R- 1 t dt = Z x Z ∞ − cos t x cos t cos t − Rdt → cos 1 − Rdt t t2 t2 1 1 1 as x → ∞. The latter improper Riemann integral exists since |t−2 cos t| ≤ t−2 and the is integrable. It follows that the improper Riemann R ∞latter cos t integral R- 1 t2 dt exists. On the other hand, on each interval [kπ, (k + 1)π ), we have | f (t)| ≥ |sin(t)|((k + 1)π )−1 . It thus follows that n Z [1,∞) 1 ∑ (k + 1)π [kπ,(k+1)π) |sin(t)| dλ(t) k =1 Z π 1 n 1 = · R- |sin(t)| dt. π k∑ k+1 0 =1 Z f dλ ≥ Since the harmonic series diverges, it follows that f is not integrable on [1, ∞). Remark 3.84. In what follows, we will also use the ‘differential’ dt in Lebesgue integrals instead of the (formally correct) dλ. We will thus write Z Z b [ a,b] f (t) dt or a f (t) dt to denote the Lebesgue integral of f on the interval [ a, b]. This is particularly helpful when the function f depends on more than one variable. To have R this feature also at hand R for general measures, we will frequently write Ω f ( x ) dµ( x ) instead of Ω f dµ to emphasize that we are integrating with respect to the variable x. 83 3 Measure and Integration 3.9 Integrals depending on a parameter Instead of [0, 1] we could use a general metric space here. The main topic of this section is to interchange operations like integration, differentiation and taking limits. This is a topic at the very heart of analysis. Suppose that (Ω, Σ, µ) is a measure space. If we are given a map f : [0, 1] × Ω →RC such that f (t, ·) is integrable for all t ∈ [0, 1], we may define F (t) := Ω f (t, x ) dµ( x ). It is then natural to ask how F depends on the parameter t ∈ [0, 1]. In this short section, we use the dominated convergence theorem to prove some results in this direction. Proposition 3.85. Let (Ω, Σ, µ) be a measure space. Furthermore, let f : [0, 1] × Ω → K be such that the following three properties hold. (a) x 7→ f (t, x ) ∈ L 1 (Ω, Σ, µ) for all t ∈ [0, 1]. (b) t 7→ f (t, x ) is continuous for almost all x ∈ Ω. (c) There exists a g ∈ L 1 (Ω, Σ, µ) such that | f (t, x )| ≤ g( x ) for all (t, x ) ∈ [0, 1] × Ω. R Then F : [0, 1] → C defined by F (t) = Ω f (t, x ) dµ( x ) is continuous. Proof. Let tn → t in [0, 1]. Then f (tn , x ) → f (t, x ) for almost all x ∈ Ω by (b). Since | f (tn , x )| ≤ g( x ) for all x ∈ Ω by assumption and g ∈ L 1 (Ω), it follows from the dominated convergence theorem, Theorem 3.75, that F (tn ) = Z Ω f (tn , x ) dµ( x ) → Z Ω f (t, x ) dµ( x ) = F (t). This proves the continuity of F. Proposition 3.86. Let I be an interval in R and (Ω, Σ, µ) be a measure space. Furthermore, let f : I × Ω → K be such that the following three properties hold. (a) x 7→ f (t, x ) ∈ L 1 (Ω, Σ, µ) for all t ∈ I. (b) t 7→ f (t, x ) is differentiable for all x ∈ Ω. (c) There exists a g ∈ L 1 (Ω, Σ, µ) such that | ∂t∂ f (t, x )| ≤ g( x ) for all (t, x ) ∈ I × Ω. R Then F : I → K defined by F (t) = Ω f (t, x ) dµ( x ) is differentiable. Moreover, ∂ ∂t f ( t, x ) is integrable for all t ∈ I and F 0 (t) = d dt Z Ω f (t, x ) dµ( x ) = Z Ω ∂ f (t, x ) dµ( x ). ∂t Proof. Fix t ∈ I and let (tn ) be a sequence in I that converges to t. Define hn , h : Ω → C by hn ( x ) := (tn − t)−1 ( f (tn , x ) − f (t, x )) and h( x ) = ∂ ∂t f ( t, x ). Then hn is integrable for every n ∈ N as a linear combination of integrable functions. Moreover, hn ( x ) → h( x ) for all x ∈ Ω by assumption. By the mean-value theorem, hn ( x ) = ∂t∂ f (ξ n , x ) for some ξ n between 84 3.10 Product measures t and tn . In particular, |hn | ≤ g. Thus the dominated convergence theorem shows that h is integrable and F (tn ) − F (t) = tn − t Z Ω hn ( x ) dµ( x ) → Z Ω h( x ) dµ( x ) = Z Ω ∂ f (t, x ) dµ( x ). ∂t This finishes the proof. 3.10 Product measures In this section we construct a σ-algebra and a corresponding measure on the product of two suitable measure spaces. Our motivation is to extend the theory in order to deal with iterated integrals. The product measure will allow us to write an iterated integral as a single integral with resprect to the product measure. Definition 3.87. Let (Ωk , Σk ) be a measurable space for k = 1, . . . , n. The product (measurable space) of the spaces (Ωk , Σk ) is the measurable N space (∏nk=1 Ωk , nk=1 Σk ), where ∏nk=1 Ωi is the Cartesian product of the N sets Ωk , i.e., the set of all tuples ( x1 , . . . , xn ) where xk ∈ Ωk and nk=1 Σk is generated by the cuboids A1 × · · · × An where Ak ∈ Σk . Exercise 3.88. Let (Ω, Σ) and, for k = 1, . . . , n, also (Ωk , Σk ) be measure spaces. Let f k : Ω → Ωk be a function and define f : Ω → ∏nk=1 Ωk by N f ( x ) = ( f 1 ( x ), . . . , f n ( x )). Show that f is Σ/ nk=1 Σk -measurable if and only if f k is Σ/Σk -measurable for all k = 1, . . . , n. In the following, let (Ωi , Σi , µi ) be σ-finite measure spaces for i = 1, 2. We define a measure µ1 ⊗ µ2 on the σ-algebra Σ1 ⊗ Σ2 which is the product of the measures µ1 and µ2 in the sense that µ1 ⊗ µ2 ( A × B ) = µ1 ( A ) µ2 ( B ) for all A ∈ Σ1 and B ∈ Σ2 . Note that as µ1 and µ2 are σ-finite, by Corollary 3.38 there exists at most one such measure. For a set Q ⊂ Ω1 × Ω2 and x ∈ Ω1 , y ∈ Ω2 , we define the cuts [ Q] x and [ Q]y by [ Q] x := {y ∈ Ω2 : ( x, y) ∈ Q} and [ Q]y := { x ∈ Ω1 : ( x, y) ∈ Q}. Lemma 3.89. For x ∈ Ω1 , y ∈ Ω2 and Q ∈ Σ1 ⊗ Σ2 we have [ Q] x ∈ Σ2 and [ Q ] y ∈ Σ1 . Proof. We put G := { Q ∈ Σ1 ⊗ Σ2 : [ Q] x ∈ Σ2 }. We claim that G is a σalgebra on Ω1 × Ω2 . Clearly (S1) holds, since [Ω1 × Ω2 ] x = Ω2 . (S2) and 85 Observe that the 2-dimensional Lebesgue measure λ2 is the product measure of the one-dimensional Lebesgue measure with itself.